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a(n) is the largest unitary divisor of sigma(n) such that its every prime factor p also divides A003961(n), and valuation(sigma(n),p) >= valuation(A003961(n),p).
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%I #11 Feb 17 2022 00:02:05

%S 1,3,1,1,1,3,1,1,1,9,1,1,1,3,1,1,1,3,1,7,1,9,1,5,1,3,1,1,1,9,1,1,1,27,

%T 1,1,1,3,1,1,1,3,1,1,1,9,1,1,1,3,1,1,1,3,1,1,5,9,1,7,1,3,1,1,7,9,1,9,

%U 1,9,1,1,1,3,1,1,1,3,1,1,1,9,1,1,1,3,5,1,1,9,1,1,1,9,1,1,1,9,13,1,1,27

%N a(n) is the largest unitary divisor of sigma(n) such that its every prime factor p also divides A003961(n), and valuation(sigma(n),p) >= valuation(A003961(n),p).

%H Antti Karttunen, <a href="/A351545/b351545.txt">Table of n, a(n) for n = 1..65537</a>

%H <a href="/index/Pri#prime_indices">Index entries for sequences computed from indices in prime factorization</a>

%H <a href="/index/Si#SIGMAN">Index entries for sequences related to sigma(n)</a>

%F a(n) = Product_{p^e || A000203(n)} p^(e*[p divides A003961(n) but p^(1+e) does not divide A003961(n)]), where [ ] is the Iverson bracket, returning 1 if the condition holds, and 0 otherwise. Here p^e is the largest power of prime p dividing sigma(n).

%F a(n) = A000203(n) / A351547(n).

%F For all n >= 1, a(n) is a divisor of A351544(n), which is a divisor of A000203(n).

%o (PARI)

%o A003961(n) = { my(f = factor(n)); for(i=1, #f~, f[i, 1] = nextprime(f[i, 1]+1)); factorback(f); };

%o A351545(n) = { my(s=sigma(n),f=factor(s),u=A003961(n)); prod(k=1,#f~,if(!(u%f[k,1]) && (f[k,2]>=valuation(u,f[k,1])), f[k,1]^f[k,2], 1)); };

%Y Cf. A000203, A003961, A351544, A351547.

%K nonn

%O 1,2

%A _Antti Karttunen_, Feb 16 2022